Theory of the Lattice Boltzmann Method

Transcript

1 Theory of the Lattce Boltzmann Method Burkhard Dünweg Max Planck Insttute for Polymer Research Ackermannweg Manz B. D. and A. J. C. Ladd, arxv: v2, Advances n Polymer Scence 221, 89 (2009)

2 Lattce Boltzmann lnearzed Boltzmann equaton (knetc theory of gases) fully dscretzed stes r, lattce spacng a tme t, tme step h c small set of veloctes c h connects two stes n ( r, t): real number, mass densty on ste r correspondng to velocty c n ( r + c h, t + h) = n ( r, t) = n ( r, t) + ( r, t)

3 Conservaton laws, symmetres n ( r + c h, t + h) = n ( r, t) = n ( r, t) + {n ( r, t)} ρ = n j = ρ u = n c = c = 0 mass conservaton momentum conservaton localty rotatonal symmetry (lattce!) Galle nvarance (fnte number of veloctes)

4 Low Mach number physcs only u c only u c s Ma = u/c s 1 low Mach number compressblty does not matter equaton of state does not matter choose deal gas! m p partcle mass: p = ρ m p k B T c 2 s = p ρ = 1 m p k B T p = ρc 2 s k B T = m p c 2 s

5 Where we want to get n the contnuum lmt a 0, h 0: t ρ + α j α = 0 t j α + β ( ρc 2 s δ αβ + ρu α u β ) = β σ αβ η αβγδ = σ αβ = η αβγδ γ u δ (ζ 23 η ) δ αβ δ γδ + η (δ αγ δ βδ + δ αδ δ βγ ) η shear vscosty ζ bulk vscosty

6 Dfference equaton dfferental equaton Example: consder f (x + a) f (x) = g(x) small a: a d f (x) g(x) dx contnuum lmt a 0: d g(x) f (x) = lm dx a 0 a watch out: f the rhs does not exst, then the contnuum lmt does not exst! correctons to the leadng behavor?

7 set f (x + a) f (x) = g(x) g ndependent of a D = a d dx g(x) = a ( g 0 (x) + ag 1 (x) + a 2 g 2 (x) +... ) f (x 0 + a) = f (x 0 ) + a df dx + a2 d 2 f x=x0 2! dx x=x0 ( = exp a d ) f (x) dx = exp(d)f (x) x=x0 x=x0 [exp(d) 1]f (x) = g(x) f (x) = [exp(d) 1] 1 g(x) Df (x) = D [exp(d) 1] 1 g(x)

8 now, x exp(x) 1 = 1 x 2 + x2 12 x x = Bernoull numbers B k k=0 B k k! xk f (x + a) f (x) = g(x) ] D2 D2 Df (x) = [ g(x) ] d [1 dx f (x) = D2 D [ g0 (x) + ag 1 (x) + a 2 g 2 (x) +... ] systematc expanson n powers of a!

9 LB contnuum lmt: how?? a 0, h 0 to be replaced by ε 0: wave lke scalng: a/h = const. a = εa 0 h = εh 0 c = c 0 n ( r + ε c 0 h 0, t + εh 0 ) n ( r, t) = dffusve scalng: a 2 /h = const. a = εa 0 h = ε 2 h 0 c = ε 1 c 0 n ( r + ε c 0 h 0, t + ε 2 h 0 ) n ( r, t) =

10 Expandng the soluton lhs s O(ε) rhs must be O(ε)! ansatz: n = n (0) + εn (1) + O(ε 2 ) {n } = (0) + ε (1) + O(ε 2 ) { } = n (0) + ε n j j =: ε j L j n (1) j + O(ε 2 ) n =n (0) n (1) j + O(ε 2 ) and conservaton laws: { (k) n (0) = } = 0 (k) c = 0

11 Zeroth order 0 = (0) = {n (0) } {n (0) } collsonal nvarant, {n (0) } = n eq no spurous conservaton laws n (0) = n (0) (ρ, j)

12 No expanson for conserved quanttes! n = n (0) + εn (1) + O(ε 2 ) ρ = ρ (0) + ερ (1) + O(ε 2 ) j = j (0) + ε j (1) + O(ε 2 ) n (0) = n (0) = n (0) ( ) ρ, j ( ρ (0) + ερ (1) + O(ε 2 ), j (0) + ε j (1) + O(ε 2 ) ) no expanson for n (0) ρ (1) = ρ (2) =... = 0 j (1) = j (2) =... = 0

13 . e. ρ (0) = ρ j (0) = j n eq = ρ n eq c = j

14 Velocty moments mass densty ρ = n momentum densty stress j α = Π αβ = n c α n c α c β 3rd moment Φ αβγ = n c α c β c γ

15 LB contnuum lmt: wave lke scalng n ( r + ε c h 0, t + εh 0 ) n ( r, t) = set D = εh 0 t + εh 0 c α α D n = [ 1 D ] 2 + D ε 1 : ε 2 : (h 0 t + h 0 c α α ) n (0) = (1) (h 0 t + h 0 c α α )n (1) = (2) 1 2 (h 0 t + h 0 c α α ) (1)

16 take zeroth and frst velocty moment: ε 1 : or ε 2 : h 0 t ρ + h 0 α j α = 0 h 0 t j β + h 0 α Π (0) αβ = 0 t ρ + α j α = 0 t j β + α Π (0) αβ = 0 or 0 = 0 h 0 α Π (1) αβ = 1 ( ) 2 h 0 α Π (1) αβ Π(1) αβ 1 ( ) 2 α Π (1) αβ + Π(1) αβ = 0

17 consequences: Π (0) depends only on ρ, j (locally!!) Π (0) must be the Euler stress!. e. n eq c α c β = ρc 2 s δ αβ + ρu α u β stress relaxaton at order ε 2 gves rse to some sort of dsspaton, but no relaton to the prevous order. e. relaton to velocty gradents (vscous shear stresses) can not be establshed! we fnd: Euler equatons at order ε 1, but no useful results beyond! can we at least adjust n eq such that we get Euler n the leadng order? YES!

18 The equlbrum populatons ansatz (Euler stress s a 2nd order polynomal n u!): n eq (ρ, u) = w ρ ( 1 + A u c + B( u c ) 2 + Cu 2) w postve weghts, dentcal wthn a shell. cubc symmetry: w = 1 w c α = 0 w c α c β = σ 2 δ αβ w c α c β c γ = 0 w c α c β c γ c δ = κ 4 δ αβγδ +σ 4 (δ αβ δ γδ + δ αγ δ βδ + δ αδ δ βγ )

19 mass: ρ = n eq = ρ ( w 1 + A u c + B( u c ) 2 + Cu 2) 0 = Bu α u β w c α c β + Cu 2 = Bu α u β σ 2 δ αβ + Cu 2 = (Bσ 2 + C) u 2 C + Bσ 2 = 0

20 momentum: ρu α = n eq c α = ρ ( w c α 1 + A u c + B( u c ) 2 + Cu 2) = ρau β w c α c β = ρau β σ 2 δ αβ = ρaσ 2 u α Aσ 2 = 1

21 stress: c 2 s δ αβ + u α u β = 1 ρ = n eq c α c β w c α c β ( 1 + A u c + B( u c ) 2 + Cu 2) hence κ 4 = 0 and = ( 1 + Cu 2) σ 2 δ αβ + Bu γ u δ κ 4 δ αβγδ + Bu γ u δ σ 4 (δ αβ δ γδ + δ αγ δ βδ + δ αδ δ βγ ) c 2 s δ αβ + u α u β = ( σ 2 + Cσ 2 u 2 + Bσ 4 u 2) δ αβ + 2Bσ 4 u α u β. e. σ 2 = c 2 s, Cσ 2 + Bσ 4 = 0, 2Bσ 4 = 1

22 taken all together: κ 4 = 0 σ 2 = cs 2 2Bσ 4 = 1 Cσ 2 + Bσ 4 = 0 C + Bσ 2 = 0 Aσ 2 = 1 sx equatons, sx unknowns. multply Eq. 5 wth σ 2 and compare wth Eq. 4. hence the soluton s: κ 4 = 0 σ 2 σ 4 = c 2 s = c 4 s A = 1/c 2 s B = 1/(2c 4 s ) C = 1/(2c 2 s )

23 form of the equlbrum populatons s ( n eq (ρ, u) = w ρ 1 + u c c 2 s + ( u c ) 2 2c 4 s ) u2 2cs 2 what are the weghts? we need to satsfy the three condtons: w = 1 κ 4 = 0 σ 4 = σ 2 2 therefore, at least three shells are needed! each shell s assgned ts own σ 2, σ 4, κ 4 (assumng weght one).

24 D3Q19 one zero velocty: c = 0, weght w 0 sx nearest neghbors: c = (a/h)(±1, 0, 0), (a/h)(0, ±1, 0), (a/h)(0, 0, ±1), weght w I twelve next nearest neghbors: c = (a/h)(±1, ±1, 0), (a/h)(±1, 0, ±1), (a/h)(0, ±1, ±1), weght w II zeroth shell: velocty moments trval

25 frst shell: c 2 1 c 4 1 c 2 1c 2 2 = 2(a/h) 2 = σ 2 (I) = 2(a/h) 4 = κ 4 (I) + 3σ 4 (I) = 0 = σ 4 (I) σ 2 (I) = 2(a/h) 2 σ 4 (I) = 0 κ 4 (I) = 2(a/h) 4

26 second shell: c 2 1 c 4 1 c 2 1c 2 2 = 8(a/h) 2 = σ 2 (II) = 8(a/h) 4 = κ 4 (II) + 3σ 4 (II) = 4(a/h) 4 = σ 4 (II) σ 2 (II) = 8(a/h) 2 σ 4 (II) = 4(a/h) 4 κ 4 (II) = 4(a/h) 4

27 0 = κ 4 = w I κ 4 (I) + w II κ 4 (II) = 2w I 4w II w I σ 2 σ 4 = 2w II = w I σ 2 (I) + w II σ 2 (II) = w II (2σ 2 (I) + σ 2 (II)) = w II (a/h) 2 ( ) = 12w II (a/h) 2 = w I σ 4 (I) + w II σ 4 (II) = w II (2σ 4 (I) + σ 4 (II)) = 4w II (a/h) 4 = σ 2 2 = 144w 2 II (a/h)4

28 1 = 36w II w II = 1 36 w I = 2w II = = w 0 + 6w I + 12w II = w w 0 = 1 3 c 2 s = σ 2 = 12w II (a/h) 2 = 1 3 (a/h)2 all coeffcents of n eq known!

29 LB contnuum lmt: dffusve scalng n ( r + ε c 0 h 0, t + ε 2 h 0 ) n ( r, t) = watch out: c = ε 1 c 0! defne moments wrt c 0, not c! e. g. j = n c 0 etc.! set: D n = D = ε 2 h 0 t + εh 0 c 0α α [ 1 D ] 2 + D ε 1 : ε 2 : h 0 c 0α α n (0) = (1) h 0 t n (0) + h 0 c 0α α n (1) = (2) 1 2 h 0c 0α α (1)

30 1st order zeroth velocty moment: frst velocty moment: h 0 c 0α α n (0) = (1) α j α = 0 α Π (0) αβ = 0

31 2nd order h 0 t n (0) + h 0 c 0α α n (1) = (2) 1 2 h 0c 0α α (1) zeroth velocty moment: ncompressble flud! 1st velocty moment: t j β + α Π (1) αβ t j β + 1 ( ) 2 α Π (1) αβ + Π(1) αβ t ρ = 0 = 1 2 α = 0 ( ) Π (1) αβ Π(1) αβ

32 Addng the equatons t ρ + α j α = 0 t j β + α Π (0) αβ + 1 ( ) 2 α Π (1) αβ + Π(1) αβ = 0 looks lke Naver Stokes; Π (0) Euler stress, (1/2) ( Π (1) + Π (1)) vscous stress; BUT dynamcs wth constrants: α j α = 0 α Π (0) αβ = 0 t ρ = 0 ncompressble pressure as a Lagrange multpler dffcult to analyze! (Junk / Luo / Klar)

33 All these dffcultes go away when one combnes wave lke and dffusve scalng n a multple tme scale analyss! So, what s ths?

34 The dea of multple tme scale expanson example: damped oscllator T oscllaton perod τ frctonal relaxaton tme d 2 dt 2x + 1 d τ dt x + 1 T 2x = 0 consder T τ (weak dampng) try to treat ε := T 2τ as a small parameter for perturbaton theory unt system: set T = 1 d 2 dt 2x + 2ε d dt x + x = 0

35 d 2 dt 2x + 2ε d dt x + x = 0 x(t = 0) = 1, ẋ(t = 0) = ε exactly solvable ( ) x(t) = exp( εt)cos 1 ε 2 t ε dependence looks harmless, but...

36 Nave perturbaton theory yelds herarchy: d 2 dt 2x + 2ε d dt x + x = 0 x(t) = x 0 (t) + εx 1 (t) + ε 2 x 2 (t) +... ẍ k + x k = 2ẋ k 1 wth (def.) x 1 = 0, plus correspondng herarchy of ntal condtons ε 0 : ε 1 : x 0 = cos t x 1 = t cos t

37 . e., 1st order perturbaton theory yelds x(t) = (1 εt)cos t + O(ε 2 ) dentcal wth Taylor expanson of the exact soluton! For t 1/ε ths becomes completely useless!!! 1.5 damped harmonc oscllator, epslon = x t

38 Defcences of nave perturbaton theory: does not capture the presence of dfferent tme scales (here: fast oscllatons vs. slow dampng) typcally, ths occurs f one has qualtatvely dfferent behavor for ε = 0 and small ε > 0 (here: conservatve vs. dsspatve) sngular perturbaton theory needed

39 Multple tme scale analyss Idea: ( ) x(t) = exp( εt)cos 1 ε 2 t exp( εt)cos t = exp( t 1 )cos t = x (t, t 1 ) wth t 1 = εt consder x as a functon of two ndependent varables t, t 1 should be able to grasp the tme scale separaton! hence, study expanson x (t, t 1 ) = x 0 (t, t 1 ) + εx 1 (t, t 1 ) + ε 2 x 2 (t, t 1 ) +... wth d dt = t + t 1 = t t 1 t + ε t 1

40 agan the damped oscllator: (expand both x and p) d dt x = p d dt p = 2εp x ε 0 : t x 0 = p 0 t p 0 = x 0 x 0 = A(t 1 )cos t + B(t 1 )snt = A(t 1 )snt + B(t 1 )cos t p 0 A(t 1 ), B(t 1 ) not yet known

41 ε 1 : ansatz t x 1 t p 1 = p 1 t 1 x 0 = x 1 t 1 p 0 2p 0 x 1 p 1 = C(t, t 1 )cos t + D(t, t 1 )snt = C(t, t 1 )snt + D(t, t 1 )cos t yelds C t D t = A t 1 2Asn 2 t + 2B snt cos t = B t 1 2Bcos 2 t + 2Asnt cos t ntegrate wrt t, but the soluton should not explode!!!

42 sn 2 t = cos 2 t = 1 2 now, hence or A t 1 + A = 0 B t 1 + B = 0 A = Âexp( t 1 ) B = ˆB exp( t 1 ) nsert ths nto ε 0 soluton, ntal condtons: x 0 = exp( εt)cos t

43 x dfference exact vs. perturbaton theory, eps = t

44 Chapman Enskog expanson orgnal LBE: n ( r + c h, t + h) n ( r, t) = desred: contnuum lmt h 0, c fxed set h = εh 0 expanson parameter ε 1, ε 0 wrte t 1 = t, r 1 = r yelds: n ( r 1 + c εh 0, t 1 + εh 0 ) n ( r 1, t 1 ) = two tme scales: waves: tme length dffuson: tme (length) 2 second tme scale: t 2 = εt

45 study LBE: n ( r 1 + ε c h 0, t 1 + εh 0, t 2 + ε 2 h 0 ) n ( r 1, t 1, t 2 ) = r = r 1 set t = t 1 + ε t 2 D = εh 0 c α 1α + εh 0 t1 + ε 2 h 0 t2 D n = [ 1 D ] 2 + D

46 ε orders ε 1 : ε 2 : (h 0 c α 1α + h 0 t1 )n (0) = (1) h 0 t2 n (0) + (h 0 c α 1α + h 0 t1 ) n (1) = (2) 1 2 (h 0c α 1α + h 0 t1 ) (1) or h 0 t2 n (0) + 1 ( ) 2 (h 0c α 1α + h 0 t1 ) n (1) + n (1) = (2)

47 Zeroth velocty moment: Mass conservaton t1 ρ + 1α j α = 0 t2 ρ = 0 Hence, contnuty equaton OK!!!

48 Frst velocty moment: Momentum conservaton t2 j α β t1 j α + 1β Π (0) αβ = 0 ( ) Π (1) αβ + Π(1) αβ = 0 comparson wth Naver Stokes: Euler stress: Π (0) αβ = ρc2 s δ αβ + ρu α u β Newtonan vscous stress: ε 2 ( Π (1) αβ + Π(1) αβ ) = σ αβ

49 Second velocty moment: A useful relaton from explct form of n eq : ( ) t1 Π (0) αβ + 1γΦ (0) αβγ = h 1 0 Π (1) αβ Π(1) αβ use contnuty and Euler for (neglectng terms O(u 3 )): (detals see next three sldes) Φ (0) αβγ = ρc2 s (u α δ βγ + u β δ αγ + u γ δ αβ ) t1 Π (0) αβ = t 1 (ρc 2 s δ αβ + ρu α u β ) =... Π (1) αβ Π(1) αβ = h 0ρc 2 s ( α u β + β u α )

50 Calculaton of Φ n eq (ρ, u) = w ρ ( 1 + u c c 2 s + ( u c ) 2 2c 4 s ) u2 2cs 2 hence Φ αβγ = ρ cs 2 u δ w c α c β c γ c δ = ρ cs 2 u δ cs 4 (δ αβ δ γδ + δ αγ δ βδ + δ αδ δ βγ ) = ρc 2 s (δ αβ u γ + δ αγ u β + δ βγ u α )

51 Equaton of moton for the Euler stress pure Euler hydrodynamcs Euler equatons: D = Dt t + u γ γ D Dt ρ = ρ γu γ ρ D Dt u α D Dt Π αβ neglect O(u 3 ): = c 2 s α ρ Π αβ = ρc 2 s δ αβ + ρu α u β = ( c 2 s δ αβ + u α u β ) D Dt ρ + u αρ D Dt u β + u β ρ D Dt u α = ρ ( c 2 s δ αβ + u α u β ) γ u γ c 2 s u α β ρ c 2 s u β α ρ t Π αβ + u γ c 2 s δ αβ γ ρ = c 2 s δ αβ ρ γ u γ c 2 s u α β ρ c 2 s u β α ρ

52 t Π αβ + c 2 s δ αβ γ (ρu γ ) + c 2 s u α β ρ + c 2 s u β α ρ = 0 t Π αβ + c 2 s δ αβ γ (ρu γ ) + c 2 s β (ρu α ) + c 2 s α (ρu β ) = c 2 s ρ β u α + c 2 s ρ α u β t Π αβ + γ { ρc 2 s (δ αβ u γ + δ βγ u α + δ αγ u β ) } = c 2 s ρ β u α + c 2 s ρ α u β t Π αβ + γ Φ αβγ = ρc 2 s ( α u β + β u α )

53 Lnear collson operator = (0) + ε (1) + O(ε 2 ) = ε (1) + O(ε 2 ) O(ε 2 ) does not contrbute to hydrodynamcs gnore (1) = n j n (1) j = L j n (1) j j {n 0 k } j. e. = j ( ) L j n j n eq j

54 The lnear collson process n neq := n n eq n = n + j L j n neq j n neq = n neq + j L j n neq j Γ j := δ j + L j n neq = j Γ j n neq j Γ =??? smplest choce: Lattce BGK: ( Γ j = 1 1 ) δ j τ study here the MRT (mult relaxaton tme) framework!

55 n neq = j 0 = j Γ j n neq j ( ) Γ j n neq j 0 = Γ j e 0 := 1 e 0j 0 = e 0 Γ j. e. e 0 s left egenvector, egenvalue zero.

56 c x n neq = j 0 = j Γ j n neq j ( ) c x Γ j n neq j 0 = c x Γ j e 1 := c x e 1j 0 = e 1 Γ j. e. e 1 s left egenvector, egenvalue zero. analogous: e 2 = c y, e 3 = c z

57 n neq = j Γ j n neq j c γc γ (bulk stress) (bulk stress relaxaton wth γ b ): Πγγ neq = ( ) c γ c γ Γ j n neq j j Π neq γγ = γ b Π neq γγ = γ b j n neq j c jγ c jγ = j (γ b c jγ c jγ ) n neq j γ b c jγ c jγ = c γ c γ Γ j e 4 := c γ c γ e 4j γ b = e 4 Γ j. e. e 4 s left egenvector, egenvalue γ b

58 ...and so on! e 5,..., e 9 : fve shear stresses, egenvalue γ s (same value for symmetry reasons) e 10,..., e 18 9 knetc modes, ghost modes (hgher order polynomals n the c ). e. we do not know Γ drectly, but ts egenvalues and egenvectors! generally (egenvalues γ ) e kj γ k = e k Γ j γ 1 for lnear stablty!

59 Modes set m k m neq k = j = = j e kj n j e k n neq = ( ) e k Γ j e k n neq j j = j Γ j n neq j e kj γ k n neq j = γ k m neq k I. e. the relaxaton process s smple n mode space! γ 0 = γ 1 = γ 2 = γ 3 = 0 (mass and momentum conservaton) γ 4 = γ b (bulk stress) γ 5 =... = γ 9 = γ s (shear stress) γ 10 =... = γ 18 = 0 (smplest choce, not necessary)

60 Orthogonalty scalar product: n n = w n n Clam: It s possble to pck the egenvectors n such a way that they satsfy e k e l = N k δ kl where the N k are just normalzaton constants. Proof: Ether you understand group theory pretty well, or you do an explct Gram Schmdt orthogonalzaton! Result s tabulated n the revew!

61 δ kl = = 1 w e k e l N k w e k N k w N l e l ê k := w N k e k δ kl = ê k ê l. e. ê k s a standard orthogonal matrx wth Eucldean scalar product!

62 m k = e k n = ˆm k := 1 m k Nk Nk w ê k n ˆn := 1 w n ˆm k = ê kˆn orthonormal transformaton, trval to nvert!

63 Lnear stress relaxaton and vscostes Π neq αβ Π αβ = Π αβ δ αβπ γγ Π neq αβ Π neq γγ = γ s Π neq αβ = γ b Π neq γγ Π neq αβ = Π neq αβ Π neq αβ δ αβ ( Π neq γγ Π neq ) γγ = (γ s 1) Π neq αβ δ αβ (γ b 1) Π neq γγ

64 on the other hand, we had derved Π neq αβ Π neq αβ = hρcs 2 ( α u β + β u α ) = hρcs ( 2 α u β + β u α 23 ) δ αβ γ u γ comparson: or (γ s 1) Π neq αβ (γ b 1) Π neq γγ hρc2 s δ αβ γ u γ = hρc 2 s = 2hρc 2 s γ u γ ( α u β + β u α 23 δ αβ γ u γ ) Π neq αβ = hρc2 s ( α u β + β u α 23 ) γ s 1 δ αβ γ u γ Π neq γγ = 2hρc2 s γ b 1 γu γ

65 Chapman Enskog told us: Newtonan vscous stress s σ αβ = 1 ( ) Π neq 2 αβ + Π neq αβ hence 1 ) ( Π neq 2 αβ + Π neq αβ = 1 2 (γ s + 1) Π neq αβ = hρc2 s γ s + 1 ( α u β + β u α 23 ) 2 γ s 1 δ αβ γ u γ = η ( α u β + β u α 23 ) δ αβ γ u γ and δ αβ ( Π neq γγ = hρc2 s 3 = ζδ αβ γ u γ + Π neq γγ γ b + 1 γ b 1 δ αβ γ u γ ) 1 1 = 2 3 δ αβ (γ b + 1) Π neq γγ

66 read off vscostes: η = hρc2 s γ s ζ = hρc2 s 1 γ s γ b 1 γ b 3 vscosty n natural unts gamma_s γ 1 postve vscostes! any vscosty values are accessble!